The correct option is A 2(A2+B2)
Let θ be angle between →A and →B
∴ Resultant of →A and →B is
P=√A2+B2+2ABcosθ ...(i)
When →B is reversed, then the angle between →A and −→B is (180∘−θ)
Resultant of →A and −→B is
Q=√A2+B2+2ABcos(180∘−θ)
Q=√A2+B2−2ABcosθ ...(ii)
Squaring and adding (i) and (ii), we get
P2+Q2=2(A2+B2)